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ira [324]
3 years ago
8

A solution contains 15 ppm of benzene. The density of the solution is 1.00 g/mL. This means that ________.

Chemistry
1 answer:
snow_tiger [21]3 years ago
6 0

Answer:

  • second choice:<em><u> 1.0 g of the solution contains 15 × 10⁻⁶ g of benzene.</u></em>

Explanation:

ppm is a unit of concentration that means parts per million. In grams that is grams of solute per one million (10⁶) grams of solution.

Then, <em>15 ppm of benzene</em> means that there are 15 grams of benzen in 1,000,000 grams of solution.

That leads to:

  • 1,000,000 g solution / 15 g benzene

Multiplying numerator and denominator by 10⁻⁶ you find:

  • 1,000,000 × 10⁻⁶ g solution / (15 × 10⁻⁶ g benzene)

Simplifying:

  • 1.0 g solution / (15 × 10⁻⁶ g benzene)

Which is read as 1.0 g of the solution contains 15 × 10⁻⁶ g of benzene, i.e. the second answer choice.

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Pls help i beg
marusya05 [52]

Answer:

The increasing order of conductivity is O< Ge< Mn.

Explanation:

Electrical conductivity is defined as the measure of the ability of a material to conduct electrical current through it. The conductivity depends on the atomic and molecular structure of the material.

Metals are good conductors because they have a structure with many electrons with weak bonds, and this allows their movement instead non-metals have between four and eight valence electrons, which lack this tendency.

The conductivity increases in the periodic table from top to bottom and from right to left.

oxygen is a nonmetal therefore it is a bad conductor.

Germanium is a metalloid whose conductivity is greater than a nonmetal and worst than a metal.

Manganese is a metal,in this case, it is a good conductor.

4 0
3 years ago
What are some examples of how matter moves through a cycle on earth ​
olganol [36]

Answer:

Ocean, lakes and rivers. Are all liquids.

Explanation:

Ocean, lakes and rivers. Are all liquids. Snow starts off as a liquid, evaporates into a gas and camoes back as snow.

4 0
3 years ago
Given the ion C2O4-2, what species would you expect to form with each of the following ions?
Ksivusya [100]

Answer:

A. K₂C₂O₄          Potassium oxalate

B. CuC₂O₄          Copper oxalate

C. Bi₂(C₂O₄)₃         Bismuth (III) oxalate

D. Pb(C₂O₄)₂         Lead (IV) oxalate

E. (NH₄)₂C₂O₄       Ammonium oxalate

F. HC₂O₄⁻             Acid oxalate

Explanation:

C₂O₄⁻²  → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺  +  C₂O₄⁻²  → K₂C₂O₄          Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄  + 2KOH  → K₂C₂O₄  +  2H₂O

B. Cu²⁺ +  C₂O₄⁻²   ⇄  CuC₂O₄  ↓

This is a precipitate.

C.  2Bi³⁺  +  3C₂O₄⁻²   ⇄  Bi₂(C₂O₄)₃  ↓

This is a precipitate.

D.  Pb⁴⁺ +  2C₂O₄⁻²   ⇄  Pb(C₂O₄)₂  ↓

This is a precipitate.

E. 2NH₄⁺  +  C₂O₄⁻²   ⇄  (NH₄)₂C₂O₄  ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺  +  C₂O₄⁻²   ⇄  HC₂O₄⁻

HC₂O₄⁻  + H₂O  ⇄  C₂O₄⁻²  +  H₃O⁺    Ka

HC₂O₄⁻  + H₂O  ⇄  H₂C₂O₄  +  OH⁻    Kb

HC₂O₄⁻  is an amphoteric compound

6 0
3 years ago
Plz someone help, really struggling
frez [133]

Answer:

22.9 Liters CO(g) needed

Explanation:

2CO(g)     +   O₂(g)    =>    2CO₂(g)

? Liters          32.65g

                 = 32.65g/32g/mol

                 =   1.02 moles O₂

Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)

∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)

Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.

∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed

___________________

*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).

6 0
2 years ago
The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C
Dmitry [639]

Answer:

-1160kj/mol

Explanation:

the reaction is exothermic because heat is released to the environment

7 0
3 years ago
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